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3 years ago

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A plane is traveling at a velocity of 90 m/s. It accelerates at a constant rate of 1.5 m/s2 until its velocity reaches 500 m/s.

What distance did the plane cover while it was accelerating?

1 answer:

katrin2010 [14]3 years ago

5 0

Answer:

The distance the plane covered while it was accelerating is 80,633.3 m

Explanation:

Given;

initial velocity of the plane, u = 90 m/s

acceleration of the plane, a = 1.5 m/s²

final velocity of the plane, v = 500 m/s

The distance covered by the plane is given as;

v² = u² + 2ad

where;

d is the distance covered by the plane;

500² = 90² + 2(1.5)d

500² - 90² = 3d

241900 = 3d

d = 241900 / 3

d = 80,633.3 m

Therefore, the distance the plane covered while it was accelerating is 80,633.3 m

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A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferre

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Explanation:

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$\Delta U=Q+W$ (1)

Where:

$\Delta U$ is the variation in the internal (thermal) energy of the system (the value we want to find)

$Q=-100J$ is the heat transferred out of the gas (that is why it is negative)

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On the other hand, the work done on the gas is given by:

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Where:

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$\Delta V=V_{f}-V_{i}$ is the variation in volume of the gas

In this case the initial volume is $V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}$ and the final volume is $V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}$ .

This means:

$\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}$ (3)

Substituting (3) in (2):

$W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})$ (4)

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Substituting (5) in (1):

$\Delta U=-100J+180J$ (6)

Finally:

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3 years ago

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son4ous [18]

Answer:

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Yanka [14]

Answer:

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2 years ago